Phenetics

Consider these taxa, characters and states:

internal atrial two temporal pedicillate
Taxon amnion legs scales blood nostrils septum fenestrations hemipenes gizzard teeth feathers wings vertebrae
perch no no yes cold no no no no no no no no yes
coelocanth no no yes cold yes yes no no no no no no yes
salamander no yes no cold yes yes no no no yes no no yes
frog no yes no cold yes yes no no no yes no no yes
turtle yes yes yes cold yes yes no no no no no no yes
human yes yes no warm yes yes no no no no no no yes
gecko yes yes yes cold yes yes yes yes no no no no yes
snake yes no yes cold yes yes yes yes no no no no yes
alligator yes yes yes cold yes yes yes no yes no no no yes
budgy yes yes no warm yes yes yes no yes no yes yes yes

The preceding matrix can be represented numerically (for convenience) as:

1 2 3 4 5 6 7 8 9 10 11 12 13
perch 0 0 0 0 0 0 0 0 0 0 0 0 0
coelocanth 0 0 0 0 1 1 0 0 0 0 0 0 0
salamander 0 1 1 0 1 1 0 0 0 1 0 0 0
frog 0 1 1 0 1 1 0 0 0 1 0 0 0
turtle 1 1 0 0 1 1 0 0 0 0 0 0 0
human 1 1 1 1 1 1 0 0 0 0 0 0 0
gecko 1 1 0 0 1 1 1 1 0 0 0 0 0
snake 1 0 0 0 1 1 1 1 0 0 0 0 0
alligator 1 1 0 0 1 1 1 0 1 0 0 0 0
budgy 1 1 1 1 1 1 1 0 1 0 1 1 0

And, then we can construct a similarity matrix as follows...

The percent similarity for any two taxa is simply the number of characters alike for the two taxa divided by the number of characters:

	perch	coelocanth	salamander	frog	turtle	human	gecko	snake	alligator	budgy
perch		85	62	62	69	54	54	62	54	23
coelocanth			77	77	85	69	69	77	69	38
salamander				100	77	77	54	46	62	46
frog					77	77	54	46	62	46
turtle						85	85	76	85	54
human							69	62	69	69
gecko								92	85	54
snake									76	46
alligator										69
budgy

All of the phenetic methods we will review begin with this matrix and begin by connecting the two most similar taxa. In this case, there is one pair that is identical (salamander, frog) so we can draw:

          100
           :
           ,==== salamander
           |
           `==== frog

Now, how do we connect the next taxon? One must caculate the similarity of each as-yet un-grouped taxon to the two groups already formed above, and also the similarity of the two groups to each other.

In the Weighted Pair Group Method of Arithmetic averages (WPGMA), the similarity between any single taxon and some group is simply the average of the all of the involved pairwise similarities. That is, for example, the similarity between perch and (salamander, frog) is

(perch:salamander+perch:frog+salamander:frog)/3, or

= (62+62+100)/3

= 75
and then, too, the similarity between (salamander, frog) and gecko is then

(salamander:gecko+frog:gecko+ salamander:frog)/3

= (54+54+100)/3

= 69
Thus we can convert the matrix above to the following with a new column and row for the Salamander+Frog group group we created above:

	perch	coelocanth	turtle	human	gecko	snake	alligator	budgy	salamander+frog
perch		85	69	54	54	62	54	23	75
coelocanth			85	69	69	77	69	38	75
turtle				85	85	76	85	54	85
human					69	62	69	69	85
gecko						92	85	54	69
snake							76	46	64
alligator								69	75
budgy									64

Now, the most similar set of taxa is (snake, gecko) and we can draw:

      92 100
       :  :
       :  ,==== salamander
       :  |
       :  `==== frog
       :
       ,======= snake 
       |
       `======= gecko

perch coelocanth turtle human alligator budgy SaFr SnGe
perch 85 69 54 54 23 75 69
coelocanth 85 69 69 38 75 69
turtle 85 85 54 85 84
human 69 69 85 74
alligator 69 75 84
budgy 64 64
SaFr 65
SnGe

For the last entry in this table we needed to calculate the average similarity across two pre-exisitng groups (salamander frog) and (snake gecko). This is then the average of the 6 pairwise similarity values:

(100 + 92 + 54 + 54 + 46 + 46) / 6 = 65

This presents us with a bit of a problem because we have two possible things we could do with the resulting similarities: either connect turtle to Salamander + Frog or connect Human to Salamander + Frog, or create two separate equally optimal solutions and continue for each:

85  92 100
 :   :  :
 :   :  ,==== salamander
 ,======|
 |   :  `==== frog
 |   :
 `=========== turtle
     :
     ,======= snake 
     |
     `======= gecko

85  92 100
 :   :  :
 :   :  ,==== salamander
 ,======|
 |   :  `==== frog
 |   :
 `=========== human
     :
     ,======= snake 
     |
     `======= gecko

If you break ties 'systematically' (sensu PAUP*), that is according to the order of appearance in the matrix, you'd get the UPGMA tree on the left if you completed this procedure. If you broke ties randomly, you might get the tree on the right here.